Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $p = \dfrac{-2q^2 + 18q}{4q^2 + 4q - 360} \div \dfrac{q^2 + 10q}{5q^2 + 60q + 100} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{-2q^2 + 18q}{4q^2 + 4q - 360} \times \dfrac{5q^2 + 60q + 100}{q^2 + 10q} $ First factor out any common factors. $p = \dfrac{-2q(q - 9)}{4(q^2 + q - 90)} \times \dfrac{5(q^2 + 12q + 20)}{q(q + 10)} $ Then factor the quadratic expressions. $p = \dfrac {-2q(q - 9)} {4(q + 10)(q - 9)} \times \dfrac {5(q + 10)(q + 2)} {q(q + 10)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac {-2q(q - 9) \times 5(q + 10)(q + 2) } { 4(q + 10)(q - 9) \times q(q + 10)} $ $p = \dfrac {-10q(q + 10)(q + 2)(q - 9)} {4q(q + 10)(q - 9)(q + 10)} $ Notice that $(q + 10)$ and $(q - 9)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {-10q\cancel{(q + 10)}(q + 2)(q - 9)} {4q\cancel{(q + 10)}(q - 9)(q + 10)} $ We are dividing by $q + 10$ , so $q + 10 \neq 0$ Therefore, $q \neq -10$ $p = \dfrac {-10q\cancel{(q + 10)}(q + 2)\cancel{(q - 9)}} {4q\cancel{(q + 10)}\cancel{(q - 9)}(q + 10)} $ We are dividing by $q - 9$ , so $q - 9 \neq 0$ Therefore, $q \neq 9$ $p = \dfrac {-10q(q + 2)} {4q(q + 10)} $ $ p = \dfrac{-5(q + 2)}{2(q + 10)}; q \neq -10; q \neq 9 $